Base | Representation |
---|---|
bin | 10110111111001101110101… |
… | …00111111110111111100001 |
3 | 20121222120221022100202101002 |
4 | 23133212322213332333201 |
5 | 23111210112343303213 |
6 | 255302343042521345 |
7 | 13435106010113426 |
oct | 1337467247767741 |
9 | 217876838322332 |
10 | 50550601150433 |
11 | 1511a43155287a |
12 | 5805066066255 |
13 | 2228b9401cb87 |
14 | c6a93ad23d4d |
15 | 5c9e0ccd1658 |
hex | 2df9ba9fefe1 |
50550601150433 has 2 divisors, whose sum is σ = 50550601150434. Its totient is φ = 50550601150432.
The previous prime is 50550601150397. The next prime is 50550601150513. The reversal of 50550601150433 is 33405110605505.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 39687064254289 + 10863536896144 = 6299767^2 + 3295988^2 .
It is a cyclic number.
It is not a de Polignac number, because 50550601150433 - 212 = 50550601146337 is a prime.
It is not a weakly prime, because it can be changed into another prime (50550601150733) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25275300575216 + 25275300575217.
It is an arithmetic number, because the mean of its divisors is an integer number (25275300575217).
Almost surely, 250550601150433 is an apocalyptic number.
It is an amenable number.
50550601150433 is a deficient number, since it is larger than the sum of its proper divisors (1).
50550601150433 is an equidigital number, since it uses as much as digits as its factorization.
50550601150433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 135000, while the sum is 38.
Adding to 50550601150433 its reverse (33405110605505), we get a palindrome (83955711755938).
The spelling of 50550601150433 in words is "fifty trillion, five hundred fifty billion, six hundred one million, one hundred fifty thousand, four hundred thirty-three".
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